91Ó°ÊÓ

The given function value is.$f\left(t\right)=cost,-\mathrm{π}/2≤t≤\mathrm{π}/2$

On the intervalwe can write the given function as

$f\left(t\right)=\left|\cos t\right|=\{\begin{array}{ll}\cos t,& 0≤t≤\frac{\mathrm{π}}{2}\\ −\cos t,& \frac{\mathrm{π}}{2}≤t≤\mathrm{π}\end{array}$

We will find the Laplace transform of given function using the following equation

$\mathcal{L}\left\{f\right(t\left)\right\}\left(s\right)=\frac{{\displaystyle {∫}_0^T{e}^{−st}f\left(t\right)dt}}{1−e^{−sT}}$

Since .$T=\mathrm{Ï€}$ So, plug $T=\mathrm{Ï€}$into above equation as:

$\mathcal{L}\left\{f\right(t\left)\right\}\left(s\right)=\frac{{\displaystyle {∫}_0^{\mathrm{π}}{e}^{−st}f\left(t\right)dt}}{1−e^{−s\mathrm{π}}}$

Let's find $I={∫}_0^{\mathrm{π}}{e}^{−st}f\left(t\right)dt.$

$\begin{array}{c}I={∫}_0^{\mathrm{Ï€}}{e}^{−st}f\left(t\right)dt\\ =\underset{I_1}{\underset{ÁåŸ}{{∫}_0^{\frac{\mathrm{Ï€}}{2}}{e}^{−st}f\left(t\right)dt}}−\underset{I_2}{\underset{ÁåŸ}{{∫}_{\frac{\mathrm{Ï€}}{2}}^{\mathrm{Ï€}}{e}^{−st}f\left(t\right)dt}}\end{array}$$….\left(1\right)$

The expression for$I_1$ is obtained as follows:

$\begin{array}{c}{I}_1={∫}_0^{\frac{\mathrm{π}}{2}}{e}^{−st}f\left(t\right)dt\\ ={∫}_0^{\frac{\mathrm{π}}{2}}{e}^{−st}\cos tdt\\ =\left|\begin{array}{cc}u=\cos t& dv=e^{−st}dt\\ du=−\sin tdt& v=−\frac{e^{−st}}{s}\end{array}\right|\\ ={[−\frac{e^{−st}\cos t}{s}]}_0^{\frac{\mathrm{π}}{2}}−{∫}_0^{\frac{\mathrm{π}}{2}}\frac{e^{−st}\sin t}{s}dt\end{array}$

Simplify further as:

$\begin{array}{c}{\text{I}}_{\text{1}}\text{=}\frac{\text{1}}{\text{s}}\text{-}\frac{\text{1}}{\text{s}}{∫}_{\text{0}}^{\frac{\text{π}}{\text{2}}}{\text{e}}^{\text{-st}}\text{sintdt}\\ \text{=}\left|\begin{array}{cc}\text{u=sint}& {\text{dv=e}}^{\text{-st}}\text{dt}\\ \text{du=costdt}& \text{v=-}\frac{{\text{e}}^{\text{-st}}}{\text{s}}\end{array}\right|\\ \text{=}\frac{\text{1}}{\text{s}}\text{-}\frac{\text{1}}{\text{s}}\left({\left[\text{-}\frac{{\text{e}}^{\text{-st}}\text{sint}}{\text{s}}\right]}_{\text{0}}^{\frac{\text{π}}{\text{2}}}\text{+}{∫}_{\text{0}}^{\frac{\text{π}}{\text{2}}}\frac{{\text{e}}^{\text{-st}}\text{cost}}{\text{s}}\text{dt}\right)\\ \text{=}\frac{\text{1}}{\text{s}}\text{-}\frac{\text{1}}{\text{s}}\left(\text{-}\frac{{\text{e}}^{\text{-}\frac{\text{π}}{\text{2}}\text{t}}}{\text{s}}\text{+}\frac{\text{1}}{\text{s}}{\text{I}}_{\text{1}}\right)\\ \text{=}\frac{\text{1}}{\text{s}}\text{+}\frac{{\text{e}}^{\text{-}\frac{\text{π}}{\text{2}}\text{t}}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{s}}^{\text{2}}}{\text{I}}_{\text{1}}\end{array}$

From the resulting equation that

$I_1=\frac{s+e^{−\frac{\mathrm{π}}{2}s}}{s^2+\text{1}}$…… (2)

Similarly, We find

$I_2=\frac{e^{−\frac{\mathrm{π}}{2}s}−e^{−\mathrm{π}s}s}{s^2+1}$…… (3)

On substituting equation (2) and (3) in equation (1) gives

$I=\frac{s(1−e^{−\mathrm{π}s})+\text{2}{e}^{−\frac{\mathrm{π}}{2}s}}{s^2+\text{1}}$

Thus the Laplace transform of given function is obtained as

$\text{L}\left\{\text{f(t)}\right\}\text{=}\frac{\text{s}\left({\text{1-e}}^{\text{-Ï€²õ}}\right){\text{+2e}}^{\text{-}\frac{\text{Ï€}}{\text{2}}\text{s}}}{\left({\text{s}}^{\text{2}}\text{+1}\right)\left({\text{1-e}}^{\text{-Ï€²õ}}\right)}$

Rewrite above equation as:

$\text{L}\left\{f\right(t\left)\right\}\left(s\right)=\frac{s}{s^2+\text{1}}+\frac{\text{2}{e}^{−\frac{\mathrm{π}}{2}s}}{(s^2+\text{1})(\text{1}−e^{−\mathrm{π}s})}$